Question: A $5$ -meter ladder is sliding down a vertical wall so the distance between the bottom of the ladder and the wall is increasing at $6$ meters per minute. At a certain instant, the top of the ladder is $3$ meters from the ground. What is the rate of change of the area formed by the ladder at that instant (in square meters per minute)? Choose 1 answer: Choose 1 answer: (Choice A) A $6$ (Choice B) B $-7$ (Choice C) C $18$ (Choice D) D $-14$
Explanation: Setting up the math Let... $a(t)$ denote the distance between the top of the ladder and the ground at time $t$, $b(t)$ denote the distance between the bottom of the ladder and the wall at time $t$, $c$ denote the length of the ladder (which is always $5$ meters), and $A(t)$ denote the area formed by the ladder at time $t$. $a(t)$ $b(t)$ $c$ $A(t)$ We are given that $c=5$ and $b'(t)=6$. We are also given that $a(t_0)=3$ for a specific time $t_0$. We want to find $A'(t_0)$. Relating the measures The sides relate to each other through the Pythagorean theorem: $\begin{aligned} \,[a(t)]^2+[b(t)]^2&=c^2 \\\\\\ [a(t)]^2+[b(t)]^2&=5^2 \end{aligned}$ We can differentiate both sides to find an expression for $a'(t)$ : $a'(t)=-\dfrac{b(t)b'(t)}{a(t)}$ The sides and the area relate to each other through the formula for right triangle area: $A(t)=\dfrac{a(t)b(t)}{2}$ We can differentiate both sides to find an expression for $A'(t)$ : $A'(t)=\dfrac{a'(t)b(t)+a(t)b'(t)}{2}$ Using the information to solve In order to find $a'(t_0)$ we need to find $b(t_0)$. Using the Pythagorean theorem and the fact that $a(t_0)=3$ and $c=5$, we can find that $b(t_0)=4$. Let's plug ${b(t_0)}={4}$, ${b'(t_0)}={6}$, and ${a(t_0)}={3}$ into the expression for $a'(t_0)$ : $\begin{aligned} a'(t_0)&=-\dfrac{{b(t_0)}{b'(t_0)}}{{a(t_0)}} \\\\ &=-\dfrac{({4})({6})}{({3})} \\\\ &=-8 \end{aligned}$ Now let's plug ${b(t_0)}={4}$, ${b'(t_0)}={6}$, ${a(t_0)}={3}$, and $C{a'(t_0)}=C{-8}$ into the expression for $A'(t_0)$ : $\begin{aligned} A'(t_0)&=\dfrac{C{a'(t_0)}{b(t_0)}+{a(t_0)}{b'(t_0)}}{2} \\\\ &=\dfrac{(C{-8})({4})+({3})({6})}{2} \\\\ &=-7 \end{aligned}$ In conclusion, the rate of change of the area formed by the ladder at that instant is $-7$ square meters per minute. Since the rate of change is negative, we know that the area is decreasing.